5/1/2016
Chapter 27 Lecture
Atomic
Physics
The Invention of the Modern Atom
Dalton model: Atom as billiard ball
The First Atomic Theorist
(Translation)
“Everything is composed of
‘atoms’, which are physically,
but
not
geometrically,
indivisible. Between atoms lies
empty
space;
atoms
are
indestructible; have always
been, and always will be, in
motion. There are an infinite
number of atoms, and kinds of
atoms, which differ in shape
and size.”
-Democritus, 350 BC
 In 1803, John Dalton proposed a billiard
ball model that viewed the atom as a
small solid sphere.
 According to Dalton, a sample of a pure
element was composed of a large number
of atoms of a single kind.
 Each atom, regardless of type, contains an
equal amount of positively and negatively
charged subcomponents so that each
atom is electrically neutral.
V1
JJ Thomson (1897)
Theorized that “cathode
rays” are actually beams
of subatomic particles.
Thomson measured the
mass of these particles
by deflecting them in an
electric field.
Thomson model: Atom as plum
pudding
 In 1897, J. J. Thomson hypothesized that
electrons were the negatively charged
components of atoms.
 In his model, an atom was similar to a
spherically shaped plum pudding.
It is now known that a
cathode ray is a beam of
electrons.
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Thomson model: Atom as plum
pudding
An atom consists of
negative
electrons
surrounded
by
a
“pudding” of positive
charge.
The positively charged
substance
exactly
balances out the charge
of the electrons and is
also massive enough to
give
the
atom
its
measured mass.
Rutherford model: Atom as planetary
system
 In 1909, Ernest Rutherford and his
postdoctoral
colleague
Hans
Geiger
hypothesized that the atom's mass was
not spread uniformly throughout the
atom, but instead concentrated in a small
region.
But Thomson noticed a large question brought up by his
results.
The knowledge at the time was the following.
1) It was known that electrons are negatively charged.
2) It was also known that atoms are (usually) net
neutral.
The question was:
“Where does all the mass of an atom come from if
the electrons have such small mass? How can the
atom still be net neutral?”
Thomson’s
measured mass of
an electron
9 × 10-31 kg
Known mass of
lightest element
(Hydrogen)
1.7 × 10-27 kg
Ernest Rutherford (1911)
British particle physicist
who fired alpha particles at
a sheet of gold foil, and
used fluorescent paper to
detect where the alpha
particles were deflected.
 Rutherford developed a new model of the
atom in which a tiny nucleus contained
nearly all of the mass of the atom and all
of its positive charge.
V3
Rutherford’s Original Setup
Rutherford’s Surprising Setup
Some of the alpha
particles were being
deflected at extreme
angles, sometimes
straight backward!
With this one
experiment, the Plum
Pudding Model was
finished.
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Predicted by Thomson’s
Plum Pudding Model
Rutherford’s results (the large angles) are in line with
the Planetary Model of the Atom, in which a dense,
positively charged nucleus is surrounded by negative,
low-mass electrons.
Actual Results:
Explained by Rutherford’s
Planetary Model
“The result of the experiment was
as if you had fired a cannonball at a
piece of tissue paper and had it
come back and hit you.” –E.R.
Rutherford's planetary model of the
atom
 To have a steady orbit, the
electron must be moving fast
enough
that
the
force
between the electron and the
positively charged nucleus
equals the mass of the
electron times the radial
acceleration of the electron.
A difficulty with the planetary model
 An accelerating electron emits electromagnetic
radiation, and this radiation has energy.
 Because the electron orbiting the nucleus is
continually accelerating, the electron-nucleus
system should continuously lose energy.
 The Rutherford model could not explain the
stability of atoms.
Spectra of low-density gases and the
need for a new model
Spectrum Tube: The Fingerprint of an
Element
The tube is filled with a particular gas
(Hydrogen, Helium, Neon, Mercury, etc)
 Observations
show
that different gases
produce different sets
of spectral lines.
 Scientists could not
explain where the
lines came from.
A voltage is applied across the tube,
accelerating electrons through the gas.
This causes some electrons to collide with
electrons in gas atoms, transferring some
energy to the atoms.
This brings the gas atom into a higher energy
state, otherwise known as an excited state.
When the atom drops back to a lower energy,
it emits a photon (which is what you see).
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The Emission Spectrum of Hydrogen
Hydrogen atoms, after gaining electric potential energy
and then emitting a photon as they return to a lower
energy. We detect these frequencies of light during the
process:
Spectra of low-density gases and the
need for a new model
 In 1885, Johann Balmer found a pattern in
the wavelengths of the visible spectral
lines
produced
by
hydrogen
and
represented it with the following equation:
R = 1.097x107 m-1
Rydberg’s constant
Spectra of low-density gases and the
need for a new model
n
3
4
5
6
7
8
9
10

655
485
433
410
396
388
383
379
SPECTRUM
400
420
440
460
480
500
520
540
560
580
600
620
640
660
R = 1.097x107 m-1
Rydberg constant
WHITEBOARD
Lyman Series
1 1 
 R  2  2 

1 n 
1
Balmer Series
 1 1 
 R  2  2 

2 n 
Paschen Series
1 1 
 R  2  2 

3 n 
1
WHITEBOARD
 Use Balmer's equation to determine the
energies in joules and in electron volts of
the possible visible photons emitted by
hydrogen atoms.
n
 (nm) E (eV)
3
655
1.897
4
485
2.561
5
433
2.868
6
410
3.035
7
-1
R = 1.097x10 m
7
396
3.136
8
388
3.201
hc
9
383
3.246
E
10
379
3.278

Niels Bohr
“The Great Dane”
Known for his dancing
ability as well as his
Nobel Prize in
Physics.
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Bohr's model of the atom: Quantized
orbits
 In 1913, Danish physicist Niels Bohr
developed a new model that explained the
line spectra and the stability of the atom.
 He kept the structure of Rutherford's
model in terms of the small nucleus and
electrons moving around it, but
imposed a restriction on the electron
orbits.
Bohr's first postulate
 The atom is made up of a small nucleus
and an orbiting electron.
 The electron can occupy only certain
orbits, called stable orbits, which are
labeled by the positive integer n.
 When in these orbits, the electron
moves around the nucleus with a
specific value for the total energy, but
does
not
radiate
electromagnetic
waves.
 All other orbits are prohibited.
Bohr's second postulate
Bohr's third postulate
 When an electron transitions from one
stable orbit to another, the atom's energy
changes (hf = Ef – Ei).
 When the energy of the atom
decreases, the atom emits a photon.
 For the atom's energy to increase, the
atom must absorb some energy, often
by absorbing a photon.
 Because the stable orbits are discrete,
the atom can radiate or absorb only
certain specific amounts of energy.
 The stable electron orbits are the orbits
where the magnitude of the electron's
rotational (angular) momentum L is given
by:
 L is quantized; it can have only specific
discrete values (h/2).
Size of the hydrogen atom
Size of the hydrogen atom
 This smallest-radius orbit is called the
Bohr radius r1.
L
FC  Fq
me  ve2 k  q  q

rn
rn2
k  q2
me  v 
rn
2
e
me  ve2 
k  e2
rn
nh
2 
Bohr’s 3rd postulate
L  r p
Angular momentum
L  r mv
L  rn  me  ve
nh
 rn  me  ve
2 
ve 
nh
2    rn  me
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Size of the hydrogen atom
k  e2
me  v 
rn
2
e
ve 
Size of the hydrogen atom
nh
2    rn  me
rn 
2


nh
k  e2
 
me  
rn
 2    rn  me 
n
Size of the hydrogen atom
rn 
n2  h2
4   2  k  e 2  me
ve 
ve 
n2  h2
4   2  k  e 2  me
r (m)
1
5.305E-11
2
2.122E-10
3
4.774E-10
Size of the hydrogen atom
nh
2    rn  me
rn 
nh
n2  h2
2  
 me
4   2  k  e 2  me
ve 
h2
 n2
4    k  Ze 2  me
2
2    k  Ze 2 1
ve 

h
n
2    k  e2 1

h
n
Size of the hydrogen atom
 where n is known as a quantum number and
must be a positive integer.
 Only certain radii represent stable electron
orbits.
 A stable radius is said to be quantized.
QUANTIZED LEVELS
0.900
0.800
0.700
0.600
rn 
0.500
0.400
0.300
h2
 n2
4   2  k  Ze 2  me
0.200
0.100
0.000
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
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ENERGY LEVELS
ENERGY LEVELS
U e  KE  U q
Ue 
k  Ze 2
Ue  
2  rn
me  ve2  k  q  q 

 
2
 rn 
me  ve2 
k  q2
rn
me  ve2 
k  Ze 2 k  Ze 2
Ue 

2  rn
rn
Ue  
k  Ze 2
rn
Ue  
k  Ze 2
Ue  
2  rn
ENERGY LEVELS
Ue  
2   2  k 2  e 4  me
h2
 Z2 
  2 
n 
rn 
k  Ze 2
h2
2
 n2
2
4    k  Ze 2  me
2   2  k 2  e 4  me
h2
 Z2 
U e  13.6eV   2 
n 
 Z2 
U e  13.6eV   2 
n 
0.0
-2.0
n
1
2
3
4
5
6
7
-6.0
-8.0
-10.0
-12.0
U (eV)
-13.57
-3.39
-1.51
-0.85
-0.54
-0.38
-0.28
-14.0
Energy states of the Bohr model
 The energy of the atom
is inversely proportional
to the square of the
quantum number n:
 Z2 
  2 
n 
ENERGY LEVELS (Z=1)
-4.0
 Z2 
U e  2.17 x10 18 J   2 
n 
h2
 n2
2
4    k  Ze  me
2
ENERGY LEVELS (Z=1)
freedom
n = ∞ (ionization energy)
-1.51 MeV
n=3
-3.39 MeV
n=2
release a
photon
-13.6 MeV
absorve a
photon
n=1
Each “level” represents
a different electric
potential energy for the
electron/nucleus
system.
Caution! Do not take these
diagrams too literally! These
lines represent energy states
for the atom.
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Bohr Model
When an electron in an atom transitions from a more
excited state (less negative energy) to a less excited state
(more negative energy), it emits a photon!!
n=2
(first excited state)
E2 = -3.4 eV
The energy of the
photon will be equal to
the energy lost by the
atom during this
transition.
E1 = -13.6 eV
n=1
(ground state)
An electron can
only occupy certain
energy states
around a nucleus.
When it transitions
between energy
states, a photon
must either be
absorbed or
emitted by the
atom.
V4
Negative Potential Energies Again!
The more negative the EPE, the closer the electron is (on average)
from the nucleus.
Since an atom (nucleus and electrons) is in a bound state, it has a
negative electric potential energy. This means that an input of energy is
required to separate the electron and nucleus to d = ∞ (zero electric
potential energy).
The amount of energy that is needed to completely free an electron
from the atom is called the ionization energy of the atom. (You are
making the atom into a positive ion by freeing an electron)
By putting energy into the system in the right amount, we can raise
the system to a higher “energy state”. This will bring the electron (on
average) further from the nucleus.
n = ∞ (ionization energy)
0 eV
n = 3 (first excited state)
-3.39 eV
n = 2 (first excited state)
-13.6 eV
n = 1 (ground state)
d≈∞
Ground State (n = 1)
Most negative EPE
First Excited State (n = 2)
Less negative EPE
Ionization Energy (n = ∞)
Zero EPE
BOHR’S PREDICTION
Conservation of Energy works! Again!
When an atom transitions to a lower energy level, it emits a photon.
hf  U F  U i
Ephoton= Efinal - Einitial
Bohr’s 3rd postulate
Ue  
2   2  k 2  e 4  me
h2
Einitial
hf  U F  U i
Bohr’s
3rd
postulate
Ephoton
Efinal
hf 
2   2  k 2  e 4  me
h2
hf 
 Z2 
  2 
n 
 Z 2  2   2  k 2  e 4  me
  2  
h2
 nF 
2   2  k 2  e 4  me
h2
 Z2 
  2 
 ni 
 Z2 Z2 
  2  2 
 nF ni 
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BOHR’S PREDICTION (Z=1)
hf 
2   2  k 2  e 4  me
h2
hf 
1


1

2   2  k 2  e 4  me
h3  c
 1
1 
  2  2 
 nF ni 
BOHR’S PREDICTION (Z=1)
2   2  k 2  e 4  me
R
h3  c
hc
R = 1.097x107 m-1
Rydberg’s constant
 1
1 
  2  2 
n
n
i 
 F
Balmer’s
prediction !
Emission and absorption of photons
Balmer’s
prediction !
 1
1 
 R   2  2 

n
n
i 
 F
1
Ultraviolet
Visible
Infrared
The larger the energy transition
made by the electron, the more
energy the emitted photon will have.
This is why the “short” transitions
of hydrogen (the Paschen series)
emit infrared light. IR waves have
long wavelengths and lowfrequencies, thus carrying less
energy.
This is why the “long” transitions of
hydrogen (the Lyman series) emit
UV light. UV waves have short
wavelengths and high-frequencies,
thus carrying more energy.
The visible portion spectrum
(Balmer series) has energies right in
the middle.
Mystery: Solved.
Careful with Joules and eV!
Use of Planck’s constant as h = 6.6 x 10-34 J*s and/or c = 3 x 108 means
that you cannot work in eV! You must then convert everything to J!
Use this as your conversion factor:
1 J = 1.6 x 10-19 eV
With quantum and atomic phenomena, you should always expect a
reasonable number of eV for an answer (0-1000 eV).
With quantum and atomic phenomena, you should always expect a
very tiny number of Joules for an answer (5 x 10-19 J).
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WHITEBOARD 1
WHITEBOARD 1
(RANDOM ATOM NOT HYDROGEN)
(RANDOM ATOM NOT HYDROGEN)
-1 eV
n=4
9eV
-1 eV
 What  can
absorbed?
4eV 2eV
-3 eV
n=3
7eV
n=4
9eV
be
-3 eV
n=3
7eV
2eV
-5 eV
n=2
2eV
-5 eV
 What 
emitted?
5eV
-10 eV
4eV 2eV
can
n=2
be
5eV
n=1
-10 eV
WHITEBOARD 1
n=1
0 eV
4eV 2eV
-3 eV
-5 eV
2eV
n=2
0
-10 eV
200
250
300
350
400
450
500
550
600
650
n=1
Once the atom has been excited to n = 4,
all of these possible transitions can occur!
0 eV
- 1 eV
- 2 eV
- 3 eV
- 4 eV
n=4
- 7 eV
- 8 eV
Which of the following photon energies
could NOT be found in the emission
spectra of this atom after it has been
excited to the n = 4 state?
n=3
(A) 1 eV (B) 2 eV (C) 3 eV
(D) 4 eV (E) 5 eV
n=2
None of these possible transitions
have a change in energy of 4 eV.
- 5 eV
- 6 eV
n=1
1.12E-18
178
5
8E-19
249
4
6.4E-19
311
2
3.2E-19
622
 What  light can be
emitted?
same
n=2
2) Which of the following transitions
will produce the photon with the
longest wavelength?
n=1
(A) n = 2 to n = 1 (B) n = 3 to n = 1
(C) n = 3 to n = 2 (D) n = 4 to n = 1
(E) n = 4 to n = 3
- 5 eV
- 6 eV
5eV
150
138
7
(A) 1 eV (B) 2 eV (C) 3 eV
(D) 4 eV (E) 5 eV
n=3
- 3 eV
- 4 eV
SPECTRUM
1
100
 (nm)
1.44E-18
n=4
- 1 eV
- 2 eV
n=3
7eV
E (J)
9
1)Which of the following photon
energies could NOT be found in the
emission spectra of this atom after it
has been excited to the
n = 4 state?
n=4
9eV
E (eV)
A hypothetical atom has four energy states as shown below.
(RANDOM ATOM NOT HYDROGEN)
-1 eV
 What  light can be
absorbed?
- 7 eV
- 8 eV
Which of the following transitions
will produce the photon with the
0 eV
n=4 longest wavelength?
- 1 eV
- 2 eV
(E)
n=3
- 3 eV
- 4 eV
- 5 eV
- 6 eV
- 7 eV
- 8 eV
(B)
(A)
(A) n = 2 to n = 1 (B) n = 3 to n = 1
(C) n = 3 to n = 2 (D) n = 4 to n = 1
n=2
(E) n = 4 to n = 3
(C)
Longest wavelength means least energy.
(D)
n=1
(Large wavelength, low frequency, less energy.)
The lowest energy photon will be emitted during the
transition that has the smallest ΔE for the electron.
Emitted Photon Energies (left to right)
1 eV, 3 eV, 6 eV, 2 eV, 5 eV, 3 eV
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- 1 eV
- 2 eV
n=3
- 3 eV
- 4 eV
- 5 eV
- 6 eV
More Quantum Phenomena!!
Which of the following transitions
will produce the photon with the
n=4 longest wavelength?
(E)
0 eV
(B)
(A) n = 2 to n = 1 (B) n = 3 to n = 1
(C) n = 3 to n = 2 (D) n = 4 to n = 1
n=2
(E) n = 4 to n = 3
(C)
Longest wavelength means least energy.
(D)
(A)
n=1
- 7 eV
- 8 eV
(Large wavelength, low frequency, less energy.)
The lowest energy photon will be emitted during the
transition that has the smallest ΔE for the electron.
Predict the Spectrum of Neon!
Now we have a new property that is quantized:
the energy states of an atom.
Atoms can only take on certain specific electric
potential energies, and nothing in between.
This is getting more interesting…
We can predict the wavelengths of the photons emitted
during the transitions 3  2, 3  1, and 2  1.
n = ∞ (ionization energy)
0 eV
-19.36 eV
n = 3 (first excited state)
-19.6 eV
n = 2 (first excited state)
n = ∞ (ionization energy)
-19.36 eV
n = 3 (first excited state)
-19.6 eV
n = 2 (first excited state)
What color(s) will we see?
What other wavelengths could we
detect if we used IR and UV detectors?
-21.5 eV
n = 1 (ground state)
Visible Spectrum
lemitted =
Ei = E f +
hc
Ei - E f
What wavelengths can we
expect to see?
n=∞
-19.36 eV
n=3
-19.6 eV
n=2
-21.5 eV
hc
lemitted
n = 1 (ground state)
Neon
n = 2 to n = 1
First excited state to Ground state
λ = 650 nm
n = 3 to n = 1
Second excited state to Ground state
λ = 580 nm
n = 3 to n = 2
Second excited state to Ground state
-21.5 eV
n=1
λ = 5,156 nm
(Infrared)
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Final Whiteboard: Absorption and
Emission
The energy level diagram below is for a hypothetical atom. A gas of
these atoms initially in the ground state is irradiated with photons
having a continuous range of energies between 7 and 10 electron
volts. One would expect photons of which of the following energies
to be emitted from the gas?
-1 eV
n=5
-3 eV
n=4
-5 eV
n=3
-10 eV
n=2
-14 eV
n=1
(A) 1, 2, and 3 eV only
(B) 4, 5, and 9 eV only
(C) 1, 3, 5, and 10 eV only
(D) 1, 5, 7, and 10 eV only
(E) Since the original photons
have a range of energies, one
would expect a range of emitted
photons with no particular
energies.
After the electrons are excited, they
fall back down to lower energy levels
During this time, photons of energy -ΔEelectron are emitted!
-1 eV
n=5
-3 eV
n=4
-5 eV
n=3
9
eV
5
eV
-10 eV
n=2
4
eV
-14 eV
(A) 1, 2, and 3 eV only
(B) 4, 5, and 9 eV only
(C) 1, 3, 5, and 10 eV only
(D) 1, 5, 7, and 10 eV only
(E) Since the original
photons have a range of
energies, one would expect a
range of emitted photons
with no particular energies.
How to think about this
The energy level diagram below is for a hypothetical atom. A gas of
these atoms initially in the ground state is irradiated with photons
having a continuous range of energies between 7 and 10 electron volts.
If the photons transferred their energy to the atom,
-1 eV
n=5
-3 eV
n=4
-5 eV
n=3
-10 eV
n=2
-14 eV
n=1
… could jump to any
state in this range!
Electrons starting
in this state…
Since the only allowed energy
level within that range is the
second excited state (-5 eV),
the electrons in the atom will
be excited to that state!
Limitations of the Bohr model
 The wavelengths of the hydrogen spectral
lines predicted by Bohr's model are in
good agreement with the observational
evidence.
 The model does not provide predictions
that account for the spectral lines
emitted by other atoms.
 The model also does not explain the
quantization of angular momentum and
energy.
n=1
Spectral analysis
 Spectra allow
scientists to
analyze the
chemical
composition of
different
materials.
De Broglie waves and Bohr's third
postulate
 For an electron wave to be stable, an
exact integer number n of electron
wavelengths must be wrapped around the
nucleus:
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De Broglie waves and Bohr's third
postulate
De Broglie waves
p  mv
p
h

Matter wave

2r
n
n    2   r
De Broglie waves and Bohr's third
postulate
p
h
p

p  mv
mv 
mvr 
L
nh
2 
hn
2   r
hn
2   r
Max Plank
Niels
Bohr
Albert
Einstein
hn
2 
Bohr’s 3rd postulate
Louis de Broglie
Erwin Schrodinger
Planetary model
?
 Incorrect
?
Nucleus
?
?
?
?
?
 Good predictions
rn 
h2
 n2
4    k  Ze 2  me
2
?
?
 Good starting point !
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h
WHITEBOARD

De Broglie wavelength
p
m = 0.15 kg
 = 2.21x10-25 nm
V = 20 m/s
 An electron moves across a 1000 V
potential difference in a cathode ray tube.
 What is the speed of the electron?
 What is its wavelength after crossing this
potential difference?
m = 9.11x10-31 kg
-
WHITEBOARD
 = 0.1233 nm
v = 18741994 m/s
V = 5.9x106 m/s
 = 0.0389 nm
Wavelength close to
visible light, we can
treat them like a wave
Davisson and Germer Experiment
ELECTRON
WAVE DUALITY
Electrons behave like particles when interacting
with matter, but matter can not interfere !
 This model did not
support the
planetary model
 The planetary
model assumes the
electron stays as a
particle the whole
time.
Evidence of wave interference
De Broglie hypothesis: electrons propagate as waves
Schrodinger Equation
Matter
Erwin Schrodinger
Classic
mechanics
Particle
Wave
Quantum
Mechanics
Wave-Particle
duality
Macroscopic
Scale
Nanoscopic




Take any atom.
Take its properties (m, q, ….)
Do the math.
The outcome …… pictures of what the atom
looks like !!!
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LINK BETWEEN PHYSICS AND
CHEMISTRY
PHYSICS
Quantum numbers
 Principal quantum number:
 Orbital angular momentum quantum
number:
 Magnetic quantum number:
QUANTUM
MECHANICS
CHEMISTRY
Pauli exclusion principle
 Each electron in an atom must have a
unique set of quantum number, and n, l,
ml, and ms.
Possible n = 2 states of an atom
 Spin magnetic quantum number:
Possible n = 2 states of an atom
The number of states and the
quantum number designation of each
state for the 3d subshell.
Group of states that
specify the n and l
numbers
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Atomic subshells from lowest to
highest energy (approximate)
The periodic table
 The periodic table of the elements is a
natural classification of the atoms based
on their chemical properties:
 Elements in the same column have
similar electron configurations for their
outer electrons and, therefore, similar
chemical properties.
PERIODIC TABLE
The uncertainty principle
 A connection exists between the
narrowness of the slit and the apparent ycomponent of the electron's momentum
once it passes through:
https://www.youtube.com/watch?v=TQKELOE9eY4
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https://www.youtube.com/watch?v=UjaAxUO6-Uw
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